# MEASURE OF LOCATION : Deciles, Percentile and Quartiles

**WEEK 3**

**SUBJECT: FURTHER MATHEMATICS**

**CLASS: S.S 1 THIRD TERM**

**TOPIC: MEASURE OF LOCATION**

**CONTENT**:

- Mean, Mode, Median
- Deciles, Percentile and Quartiles

**Mean, Mode, Median:**

A statistical quantity which is a measure of centrality of a sample or distribution is called a **measure of location.**Another term which is frequently used interchangeably is** measure of central tendency. The most frequently used measures of location or central tendency are the mean, median and mode. **

The following are the different forms of mean

- the arithmetic mean
- the weighted mean
- the geometric mean
- the harmonic mean

The **arithmetic mean** is widely use and it is common to many. Another name for it is average. I am sure you are used to this before.

**MEAN**

To find the mean of raw data like a_{1}, a_{2}, a_{3}, a_{4}, ……a_{n}. all you need to do is add all the numbers and divide by the number e.g.

= a_{1}, a_{2}, a_{3}, a_{4}, ……a_{n}

n

Which can be written as =

**The arithmetic mean from the working mean**

**Let **X_{n}+ d_{n} where d_{n} is the deviation of X_{n} from A. Then,

ԑx_{r = }ԑ(A + d_{r})

= ԑA + ԑd_{r}

ԑx_{n} = nA + ԑd_{n}

= A + d ….(1)

The constant A in equation (1) is called the working mean, or the assumed mean or the guessed mean. Similarly, if the frequencies of x_{1, }x_{2,} x_{3,} . . . x_{n, }are respectively f_{1,} f_{2,} f_{3,}. . . f_{n} then

f_{1}

= A + ….(2)

**Mean of frequency distribution **

**Example 1:**

When table of frequency is given we follow the procedure below:

**Mean from grouped data.**

To find the mean of a grouped data;

- Find the class mark of each class interval
- This is done by adding the lower and upper class limits and divide by two.
- Determine assumed mean
- Subtract the assumed mean from class mark
- Multiply with frequency
- Sum

**Example 1:**

The data below is the masses of students in a class.

M (kg) | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 |

Freq. | 8 | 10 | 14 | 10 | 8 |

Draw the O-give and then find (a) Q_{3 }(b) Q_{2 }andQ_{3}

Solution

Q_{1} =( th = 12^{1}/_{2}th = 25.5kg

Q_{2} =2( th = 25th = 34.5kg

Q_{3} =3( th = 37^{1}/_{2}th = 44.5kg

**Example 2:** Find their mean of the data below:

X |
F |
Class Mark (C) |
C – ā |
F(c- ā) |

0 – 9 | 1 | 4.5 | -40 | -40 |

10 – 19 | 4 | 14.5 | -30 | -120 |

20 – 29 | 7 | 24.5 | -20 | -140 |

30 – 39 | 12 | 34.5 | -10 | -120 |

40 – 49 | 24 | 44.5 | 0 | 0 |

50 – 59 | 26 | 54.5 | 10 | 260 |

60 – 69 | 16 | 64.5 | 20 | 320 |

70 – 79 | 6 | 74.5 | 30 | 180 |

80 – 89 | 3 | 84.5 | 40 | 120 |

90 – 99 | 1 | 94.5 | 50 | 50 |

Total |
100 |
510 |

Let the assumed mean be ā

d = c –

= 44.5 +

= 44.5 +

= 49.6

**Measures of partition**

**Quartiles**

The median of a distribution splits the data into two equally-sized groups. In the same way, the quartiles are the three values that split a data set into four equal parts. Note that the ‘middle’ quartile, also called the second quartile denoted by Q_{1}, is the median.

The upper quartile, also called the third quartile denoted by Q_{3}, describes a ‘typical’ mark for the top half of a class and the lower quartile, also called the first quartile denoted by Q_{1}, is a ‘typical’ mark for the bottom half of the class.[mediator_tech]

**Deciles**

These are the nine points that divide a distribution into ten equal parts. The term ’decile’ is used in two different contexts. It is confusing that the same word is used in both ways. When applied to a distribution (a large group of marks), there are **nine deciles, each **of which is a** mark.**

A student whose mark is below the first decile is said to be** in decile 1. **Similarly, a student whose marks is between the first and second deciles is** in decile 2, … **and a student whose marks is above the ninth decile is **in decile 10**.When applied to individual students, the term ‘decile’ is therefore a **number between 1 and 10.**

**Percentiles **

These are the 99 points that divide a distribution into 100 equal parts.

As above, the deciles and percentiles can be derived from the Ogive using the same steps as the median or quartiles e.g. the 5^{th}decile is 5/10 and the 76^{th} percentile is the ()th item. Where n is the total frequency.

For instance;

75% =

20% =

Then, we trace the required values to the graph (curve) then to the class boundaries to get the required answer.

NOTE: A very large data set is required before the extreme percentiles can be estimated with any accuracy. (The ‘random’ variability in marks is especially noticeable in the extremes of a data set.) These percentiles can be used to categorize the individuals into percentile 1, …, percentile 100.

Example 1:

Weight (kg) | 20 – 29 | 30 – 39 | 40 – 49 | 50 – 59 | 60 – 69 | 70 –79 |

No of participants | 10 | 18 | 22 | 25 | 16 | 9 |

The frequency distribution of the weight of 100 participants in a high jump competition is as shown below:

- Construct the cumulative frequency table
- Draw the cumulative frequency curve
- From the curve, estimate:
- The median
- The lower quartile
- The upper quartile
- The inter-quartile range
- The semi inter-quartile range
- 65 percentile
- 4
^{th}decile - The probability that a participant chosen at random weighs at least 60kg

**Solution: **

Class interval Class boundary Frequency Cumulative Frequency

20 – 29 19.5 – 29.5 10 10

30 – 39 29.5 – 39.5 18 28

40 – 49 39.5 – 49.5 22 50

50 – 59 49.5 – 59.5 25 75

60 – 69 59.5 – 69.5 16 91

70 – 79 69.5 – 79.5 9 100

(b.)

(c) i. From the curve, median is half way up the distribution. This is obtained by using where N is the total frequency. Median = =

Median is at point on the graph, i.e median = 49.5kg

ii. Lower quartile is one-quarter of the way up the distribution; lower quartile = = = 25

25^{th} position

Lower quartile is at point on the graph. i.e lower quartile = 37.5kg

iii. Upper quartile is three-quarters way up the distribution;

Upper quartile =

=

=

= 75^{th} position

Upper quartile is at the point on the graph. i.e Upper quartile = 59.5kg

iv. Inter-quartile range (IQR) = Upper quartile – Lower quartile

=

= 59.5kg – 37.5kg

= 22kg

v. Semi inter-quartile range (SIQR) =

=

SIQR = 11kg

vi. 65 percentile =

=

= 65^{th} position

65 percentile is at point p on the graph = 54.5kg

vii. 4^{th}deciles =

=

= 40^{th} position

4^{th}deciles is at point d on the graph i.e 44.5kg

viii. Probability of at least 60kg = =

**CLASS ACTIVITY**

- The marks of twenty students in school are given below:

M | 1 – 3 | 4 – 6 | 7 – 9 | 10 – 12 | 13 – 15 | 16 – 18 |

F | 2 | 3 | 5 | 5 | 3 | 2 |

Draw the Ogive and from your graph. Find (i) Q_{1 }(ii) Q_{2} (iii) Q_{3 }(iv) 4^{th}decile and (v) 65^{th} percentile

- The marks scored by 30 students in a particular subject are as follows:

39, 31, 50, 18, 51, 63, 10, 34, 42, 89

73, 11, 33, 31, 41, 25, 76, 13, 26, 23

29, 30, 51, 91, 37, 64, 19, 86, 9, 20

- Prepare a frequency table using class interval of 1 – 20, 21 – 40,…
- Use your table to
- Find the mean, median
- Draw the Ogive and Histogram
- Find the value of Q
_{1}, Q_{3}of the distribution

**PRACTICE QUESTIONS**

- The table gives the distribution of marks of 60 candidates in a test.

Marks | 23-25 | 26-28 | 29-31 | 32-34 | 35 -37 | 38 –40 |

No. of students | 3 | 50 | 40 | 60 | 100 | 100 |

- Draw a cumulative frequency curve of the distribution.
- From your curve, estimate the
- 80
^{th}percentile; - Median;
- Semi-interquartile.
- The following in a data represents the scores of students in a Mathematics mock examination:
- Prepare a grouped cumulative frequency table with class intervals 20-29, 30-39, 40-49 and so on.
- Draw a cumulative frequency curve for the distribution.[mediator_tech]
- Use your curve to estimate the:
- Fail mark if 35% of the students failed the examination;
- Interquartile range.
- The table below shows the frequency distribution of the marks of 800 candidates in an examination.

Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |

Frequency | 10 | 40 | 80 | 140 | 170 | 130 | 100 | 70 | 40 | 20 |

(a) (i) Construct a cumulative frequency

table.

(ii) Draw the ogive

(iii) Use your ogive to determine the 50^{th} percentile.

- The candidate that scored less than 25% are to be withdrawn from institution, while those that scored more than 75% are to be awarded scholarship. Estimate the number of candidates that will be retained, but who will not enjoy the award.
- The following is the record of marks of 40 candidates in an examination.

65 84 91 58 43 86 73 33 76 80

57 33 53 29 40 27 72 19 51 67

37 14 18 92 13 45 61 39 23 22

22 41 27 51 63 47 19 35 39 76

(a) Using a class-interval of 11 – 20, 21 – 30 prepare

(i) a frequency table

(ii) a cumulative frequency table for distribution.

(b) Draw a cumulative frequency curve and use it to find;

(i) the median;

(ii) the lower quartile

- The following is the frequency distribution table of the marks scored by candidates in an examination.

Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |

Frequency | 2 | 7 | 8 | 13 | 24 | 30 | 6 | 5 | 3 | 2 |

(a) Make a cumulative frequency distribution use it to draw the cumulative frequency curve for the distribution.

(b) Use your graph to estimate

(i) the median mark

(ii) Lower quartile

(iii) the pass mark, if 40% of the candidate passed.

**ASSIGNMENT**

- The table below shows the weekly profit in naira from a Mini-market.

Weekly Profit (N) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |

Frequency | 1 | 3 | 2 | 0 | 1 | 6 |

(a) Draw the cumulative frequency graph of the data.

(b) From your graph, estimate the:

(i) median;

(ii) 80^{th} percent

(c) What is the modal weekly profit?

- The table below shows the frequency distribution of the marks scored by fifty students in an examination.

Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |

Frequency | 2 | 3 | 4 | 6 | 13 | 10 | 5 | 3 | 2 | 3 |

(a) Draw a cumulative frequency curve for the distribution.

(b) Use your curve to estimate the:

(i) Upper quartile;[mediator_tech][mediator_tech]

(ii) Pass mark if 60% of the students passed.

- The table below gives the ages, to the nearest 5 years, of 50 people.

Ages in years | 10 | 15 | 20 | 25 | 30 |

Numbers of people | 8 | 19 | 10 | 7 | 6 |

- Construct a cumulative frequency table for the distribution.
- Draw a cumulative frequency curve (ogive) From your ogive, find the:
- median age;
- number of people who are at most 15 years of age;
- number of people who are between 20 years and 25 years of age.

- The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.

Marks % | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 |

No of pupils | 3 | 4 | 6 | 13 | 10 | 5 | 3 | 2 |

(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.

(b) Use your curve to estimate the:

(i) cut off mark, if 300 candidates are to be offered admission.

(ii) probability that a candidate picked at random, scored at least 45%

- The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.

**Marks (in %) Frequency**

44 – 46 2

47 – 49 5

50 – 52 11

53 – 55 20

56 – 58 26

59 – 61 42

62 – 64 46

65 – 67 36

68 – 70 9

71 – 73 3

(a) Make a cumulative frequency for the distribution.

(b) Draw the cumulative frequency curve

(c) From your graph estimate the median mark

(d) The cut-off mark was 63%. What percentage of the candidates was selected?

- The table below shows the number of eggs laid by chicken in a man’s farm in a year.

No. of Eggs Per Year No. of chicken

45 – 49 10

50 – 54 36

55 – 59 64

60 – 64 52

65 – 69 28

70 – 74 10

(a) Draw a cumulative frequency curve for the distribution.

(b) Use your graph to find the interquartile range.

(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?

- The frequency distribution shows the marks of 100 students in a Mathematics test.

Mark | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |

No of Students | 2 | 4 | 9 | 13 | 18 | 32 | 13 | 5 | 3 | 1 |

(a) Draw a cumulative frequency curve for the distribution.

(b) Use your curve to estimate:

(i) the median;

(ii) the lower quartile;

(iv) the 60^{th} percentile.